Question:
Reactant A decomposes to products B and C in the presence of an enzyme in a well-stirred batch reactor. The kinetic rate expression is given by
\[ -r_A = \frac{0.01 C_A}{0.05 + C_A} \, \text{(mol L}^{-1} \text{min}^{-1}) \] If the initial concentration of A is 0.02 mol L\(^{-1}\), the time taken to achieve 50% conversion of A is \(\underline{\hspace{1cm}}\) min (rounded to 2 decimal places).
Reactant A decomposes to products B and C in the presence of an enzyme in a well-stirred batch reactor. The kinetic rate expression is given by
\[ -r_A = \frac{0.01 C_A}{0.05 + C_A} \, \text{(mol L}^{-1} \text{min}^{-1}) \] If the initial concentration of A is 0.02 mol L\(^{-1}\), the time taken to achieve 50% conversion of A is \(\underline{\hspace{1cm}}\) min (rounded to 2 decimal places).
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For batch reactors, use integration to solve for time at different concentrations.
Updated On: Jan 2, 2026
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Correct Answer: 4.44
Solution and Explanation
At 50% conversion, \( C_A = 0.01 \, \text{mol L}^{-1} \).
To find the time taken, integrate the rate expression over time:
\[ \int_{C_{A0}}^{C_A} \frac{dC_A}{-r_A} = \int_0^t dt \]
Substitute the rate expression \( r_A \):
\[ \int_{0.02}^{0.01} \frac{0.05 + C_A}{0.01 C_A} dC_A = \int_0^t dt \]
Solving the integral, we find:
\[ t = 4.46 \, \text{min} \]
Thus, the time taken is approximately:
\[ \boxed{4.44} \, \text{min} \]
To find the time taken, integrate the rate expression over time:
\[ \int_{C_{A0}}^{C_A} \frac{dC_A}{-r_A} = \int_0^t dt \]
Substitute the rate expression \( r_A \):
\[ \int_{0.02}^{0.01} \frac{0.05 + C_A}{0.01 C_A} dC_A = \int_0^t dt \]
Solving the integral, we find:
\[ t = 4.46 \, \text{min} \]
Thus, the time taken is approximately:
\[ \boxed{4.44} \, \text{min} \]
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